An aluminum link is subjected to axial tension.The loading is limited by the conditions
of not yielding and not elongating excessively. The yield limit is satisfied if the stress remains
below 50% of the 325 MPa yield stress. The elongation must remain below 0.1 mm. The link
cross-section is rectangular, 3 mm by 7 mm. Assuming elastic behavior, determine the maximum
allowable tensile force for two cases: (a) the length of the link is 35 mm and (b) the length of the
link is 55 mm.
Solution
Area of the cross section = 7*3*10^-6 = 21*10^-6 m^2
Maximum allowable stress = 325 Mpa
Allowable tensile load (T) = Stress * Area = 325*10^6 * 21*10^-6 = 6825 N
a) Young\'s modulous of alluminium is 69 Gpa
So, strain = 0.1/35 = 2.85 *10^-3
Stress = young\'s modulous * strain = 69*10^9 * 2.85*10^-3 = 197.14*10^6 = 197.14 MPa
Hence, stress is less than 325 Mpa, so the tensile load is limited by the maximum elongation
condtion.
Maximum allowed force = Stress * area = 197.14*10^6 * 21*10^-6 = 4139.9 N
b) Strain = (0.1/55) = 1.818 *10^-3
Stress = young\'s modulous * strain = 69*10^9 * 1.818*10^-3 = 197.14*10^6 = 125.44 MPa
Again stress is less than 325 Mpa, so the tensile load is limited by the maximum elongation
condtion.
Maximum allowed force = Stress * area = 125.44*10^6 * 21*10^-6 = 2634.28 N
I hope you got this :)
If any doubt then please comment.
of not yielding and not elongating excessively. The yield limit is satisfied if the stress remains
below 50% of the 325 MPa yield stress. The elongation must remain below 0.1 mm. The link
cross-section is rectangular, 3 mm by 7 mm. Assuming elastic behavior, determine the maximum
allowable tensile force for two cases: (a) the length of the link is 35 mm and (b) the length of the
link is 55 mm.
Solution
Area of the cross section = 7*3*10^-6 = 21*10^-6 m^2
Maximum allowable stress = 325 Mpa
Allowable tensile load (T) = Stress * Area = 325*10^6 * 21*10^-6 = 6825 N
a) Young\'s modulous of alluminium is 69 Gpa
So, strain = 0.1/35 = 2.85 *10^-3
Stress = young\'s modulous * strain = 69*10^9 * 2.85*10^-3 = 197.14*10^6 = 197.14 MPa
Hence, stress is less than 325 Mpa, so the tensile load is limited by the maximum elongation
condtion.
Maximum allowed force = Stress * area = 197.14*10^6 * 21*10^-6 = 4139.9 N
b) Strain = (0.1/55) = 1.818 *10^-3
Stress = young\'s modulous * strain = 69*10^9 * 1.818*10^-3 = 197.14*10^6 = 125.44 MPa
Again stress is less than 325 Mpa, so the tensile load is limited by the maximum elongation
condtion.
Maximum allowed force = Stress * area = 125.44*10^6 * 21*10^-6 = 2634.28 N
I hope you got this :)
If any doubt then please comment.
